Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{z^2 - 12z + 35}{-2z - 14} \times \dfrac{5z + 35}{9z - 63} $
First factor the quadratic. $t = \dfrac{(z - 7)(z - 5)}{-2z - 14} \times \dfrac{5z + 35}{9z - 63} $ Then factor out any other terms. $t = \dfrac{(z - 7)(z - 5)}{-2(z + 7)} \times \dfrac{5(z + 7)}{9(z - 7)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (z - 7)(z - 5) \times 5(z + 7) } { -2(z + 7) \times 9(z - 7) } $ $t = \dfrac{ 5(z - 7)(z - 5)(z + 7)}{ -18(z + 7)(z - 7)} $ Notice that $(z + 7)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ 5\cancel{(z - 7)}(z - 5)(z + 7)}{ -18(z + 7)\cancel{(z - 7)}} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $t = \dfrac{ 5\cancel{(z - 7)}(z - 5)\cancel{(z + 7)}}{ -18\cancel{(z + 7)}\cancel{(z - 7)}} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $t = \dfrac{5(z - 5)}{-18} $ $t = \dfrac{-5(z - 5)}{18} ; \space z \neq 7 ; \space z \neq -7 $